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Std 7 Mathematics · Chapter 13

13. Pythagoras’ Theorem

Step-by-step solutions to Practice Sets 48 and 49. Maharashtra State Board (Balbharati), Grade Seven Mathematics.

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Pythagoras’ theorem: in a right-angled triangle, (hypotenuse)2 = (base)2 + (height)2. The hypotenuse is always the side opposite the right angle.

Practice Set 48

1In the figures below, find the value of ‘x’.

(i) In ΔLMN, ∠M = 90°, LM = 7, MN = 24, LN = x

LN is the hypotenuse.
x2 = 72 + 242 = 49 + 576 = 625 = 252
∴ x = 25

(ii) In ΔPQR, ∠Q = 90°, PQ = 9, PR = 41, QR = x

PR is the hypotenuse.
412 = 92 + x2
1681 = 81 + x2
x2 = 1681 − 81 = 1600 = 402
∴ x = 40

(iii) In ΔEDF, ∠D = 90°, DF = 8, EF = 17, ED = x

EF is the hypotenuse.
172 = x2 + 82
289 = x2 + 64
x2 = 289 − 64 = 225 = 152
∴ x = 15

2In the right-angled ΔPQR, ∠P = 90°. If l(PQ) = 24 cm and l(PR) = 10 cm, find the length of seg QR.

∠P = 90°, so QR (opposite ∠P) is the hypotenuse.

l(QR)2 = l(PQ)2 + l(PR)2
= 242 + 102 = 576 + 100 = 676 = 262
∴ l(QR) = 26

Length of seg QR = 26 cm

3In the right-angled ΔLMN, ∠M = 90°. If l(LM) = 12 cm and l(LN) = 20 cm, find the length of seg MN.

∠M = 90°, so LN is the hypotenuse.

l(LN)2 = l(LM)2 + l(MN)2
202 = 122 + l(MN)2
400 = 144 + l(MN)2
l(MN)2 = 256 = 162
∴ l(MN) = 16

Length of seg MN = 16 cm

4The top of a ladder of length 15 m reaches a window 9 m above the ground. What is the distance between the base of the wall and that of the ladder?
9 m15 m? mwall

The wall, the ground and the ladder form a right-angled triangle; the ladder is the hypotenuse.

Let the distance be x m.
152 = 92 + x2
225 = 81 + x2
x2 = 144 = 122
∴ x = 12

The base of the ladder is 12 m from the base of the wall.

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Practice Set 49

1Find the Pythagorean triplets from among the following sets of numbers. (i) 3, 4, 5 (ii) 2, 4, 5 (iii) 4, 5, 6 (iv) 2, 6, 7 (v) 9, 40, 41 (vi) 4, 7, 8

Square the two smaller numbers and add; compare with the square of the largest number.

  • (i) 3, 4, 5 — Pythagorean triplet
    32 + 42 = 9 + 16 = 25; 52 = 25 — equal
  • (ii) 2, 4, 5 — Not a triplet
    22 + 42 = 4 + 16 = 20; 52 = 25 — not equal
  • (iii) 4, 5, 6 — Not a triplet
    42 + 52 = 16 + 25 = 41; 62 = 36 — not equal
  • (iv) 2, 6, 7 — Not a triplet
    22 + 62 = 4 + 36 = 40; 72 = 49 — not equal
  • (v) 9, 40, 41 — Pythagorean triplet
    92 + 402 = 81 + 1600 = 1681; 412 = 1681 — equal
  • (vi) 4, 7, 8 — Not a triplet
    42 + 72 = 16 + 49 = 65; 82 = 64 — not equal

The Pythagorean triplets are (i) 3, 4, 5 and (v) 9, 40, 41.

2The sides of some triangles are given below. Find out which ones are right-angled triangles? (i) 8, 15, 17 (ii) 11, 12, 15 (iii) 11, 60, 61 (iv) 1.5, 1.6, 1.7 (v) 40, 20, 30

A triangle is right-angled when the square of its longest side equals the sum of the squares of the other two sides.

  • (i) 8, 15, 17 — Right-angled
    82 + 152 = 64 + 225 = 289; 172 = 289 — equal
  • (ii) 11, 12, 15 — Not right-angled
    112 + 122 = 121 + 144 = 265; 152 = 225 — not equal
  • (iii) 11, 60, 61 — Right-angled
    112 + 602 = 121 + 3600 = 3721; 612 = 3721 — equal
  • (iv) 1.5, 1.6, 1.7 — Not right-angled
    1.52 + 1.62 = 2.25 + 2.56 = 4.81; 1.72 = 2.89 — not equal
  • (v) 40, 20, 30 — Not right-angled
    202 + 302 = 400 + 900 = 1300; 402 = 1600 — not equal

The right-angled triangles are (i) 8, 15, 17 and (iii) 11, 60, 61.

In (v) the longest side is 40, so compare 402 with 202 + 302, whatever order the sides are written in.

← 12. Perimeter and Area 14. Algebraic Formulae – Expansion of Squares →